Video: Differentiation of Logarithmic Functions

In this video, we will learn how to find the derivatives of logarithmic functions.

16:44

Video Transcript

In this video, we’ll learn how to find the derivative of the natural logarithmic function. We’ll begin by recalling the properties of logarithms that we’re going to use in this video before looking at how these properties can help us to find the derivative of simple logarithmic functions. We’ll then consider how we can apply these processes in conjunction with the standard rules for differentiation to find the derivative of more complicated logarithmic functions. Let’s begin by recalling one of the key properties of logarithms that we’re going to be using throughout this video. That is, log base 𝑎 of 𝑥 to the power of 𝑦 is the same as 𝑦 times log base 𝑎 of 𝑥.

Since the natural logarithm is just log base 𝑒, then we can say that the natural logarithm of 𝑥 to the power of 𝑦 is the same as 𝑦 times the natural logarithm of 𝑥. We’ll also lean heavily on the use of the chain rule, product rule, and quotient rule. The chain rule says that if 𝑦 is a function in 𝑢 and 𝑢 itself is a function in 𝑥, then d𝑦 by d𝑥 is equal to d𝑦 by d𝑢 times d𝑢 by d𝑥. The product rule says that, for two differentiable functions 𝑢 and 𝑣, the derivative of their product 𝑢𝑣 is equal to 𝑢 times d𝑣 by d𝑥 plus 𝑣 times d𝑢 by d𝑥. And finally, we recall the quotient rule. This says the derivative of the quotient of two differentiable functions 𝑢 and 𝑣 is 𝑣 times d𝑢 by d𝑥 minus 𝑢 times d𝑣 by d𝑥 all over 𝑣 squared.

If 𝑦 is equal to the natural log of 𝑥? What is d𝑦 by d𝑥?

There are two ways we can find the derivative of the natural log of 𝑥. The first is using differentiation from first principles. Alternatively, we can recall the following facts about functions that are inverses of one another. If 𝑓 of 𝑥 and 𝑔 of 𝑥 are inverses of each other, then we can say that 𝑔 prime of 𝑥, the derivative of our function 𝑔, is equal to one over 𝑓 prime of 𝑔 of 𝑥. This is really useful as we know the exponential function 𝑒 to the power of 𝑥 and the natural logarithmic function are inverses of one another. We can also quote the general result that the derivative of 𝑒 to the power of 𝑥 is simply 𝑒 to the power of 𝑥. So let’s let 𝑓 of 𝑥 be equal to 𝑒 to the power of 𝑥 and 𝑔 of 𝑥 be equal to the natural log of 𝑥, then 𝑔 prime of 𝑥 is equal to one over 𝑓 prime of 𝑔 of 𝑥.

We saw that 𝑓 prime of 𝑥 was equal to 𝑒 the power of 𝑥. So this means that 𝑓 prime of 𝑔 of 𝑥 is equal to 𝑒 to the power of 𝑔 of 𝑥, which is equal to 𝑒 to the power of the natural log of 𝑥. Well, 𝑒 to the power of the natural log of 𝑥 is simply 𝑥. So we obtain 𝑔 prime of 𝑥 to be equal to one over 𝑥. This means the derivative of the natural logarithm of 𝑥 is one over 𝑥 where 𝑥 is greater than zero. Now, this requirement on 𝑥 is necessary for two reasons. Firstly, 𝑥 has to be greater than zero for the natural logarithm of 𝑥. So the same must be true for its derivative. But we also know that if 𝑥 is equal to zero, then our derivative is one over zero which is undefined. We could actually redefine this and say that the derivative of the natural logarithms of the absolute value of 𝑥 is one over 𝑥 when 𝑥 is not equal to zero. We’ll be quoting this general result throughout the remainder of this video.

Find d𝑦 by d𝑥, given that 𝑦 is equal to the natural log of two 𝑥 plus seven cubed.

In this question, we have a function of a function, in other words, a composite function. We can use the chain rule to find the derivative of a composite function, though since we’re working with the natural logarithm, it’s sensible to first consider whether there is anything we can do to manipulate our expression before differentiating. We recall the general result for the natural logarithm of a power. The natural logarithm of 𝑥 to the 𝑦th power is 𝑦 times the natural logarithm of 𝑥. And we see that we can now rewrite our equation as 𝑦 equals three times the natural logarithm of two 𝑥 plus seven. This is great because we know that the derivative of a constant multiple of an expression in 𝑥 is equal to the multiple of the derivative of that expression. In other words, the derivative of three times the natural log of two 𝑥 plus seven is equal to three times the derivative of the natural log of two 𝑥 plus seven.

But how do we differentiate the natural log of two 𝑥 plus seven? We’re going to quote the general result for the derivative of the natural logarithm and we’re going to use the chain rule. The chain rule says that if 𝑦 is a function in 𝑢 and 𝑢 itself is a function in 𝑥, then the derivative of 𝑦 with respect to 𝑥 is equal to d𝑦 by d𝑢 times d𝑢 by d𝑥. And we also know that the derivative of the natural log of 𝑥 is simply one over 𝑥. So we let 𝑢 be equal to two 𝑥 plus seven such that 𝑦 is equal to the natural log of 𝑢. We know that d𝑢 by d𝑥 is two and d𝑦 by d𝑢 is one over 𝑢. The derivative then of the natural log of two 𝑥 plus seven is the product of these. It’s two times one over 𝑢. But we replace 𝑢 with two 𝑥 plus seven.

And we see that the derivative of the natural log of two 𝑥 plus seven is two over two 𝑥 plus seven. We recall we said that the derivative of three times the natural log of two 𝑥 plus seven is three times the derivative of the natural log of two 𝑥 plus seven. So the derivative is three times two over two 𝑥 plus seven, which is simply six over two 𝑥 plus seven. Now, there is actually a further standard result that we can extract from this example. That is, if 𝑦 is equal to the natural log of some function in 𝑥, then d𝑦 by d𝑥 is equal to 𝑓 prime of 𝑥, the derivative of that function, over the function itself, 𝑓 of 𝑥. If we go back to our example, we were trying to find the derivative of the natural log of two 𝑥 plus seven. We obtained that as being two over two 𝑥 plus seven. That’s the derivative of two 𝑥 plus seven over two 𝑥 plus seven. Let’s consider another example.

If 𝑓 of 𝑥 is equal to three times the natural log of two 𝑥 plus four times the natural log of 𝑥, find 𝑓 prime of one.

Here, we have a composite function. And that means we can use the chain rule to find 𝑓 prime of 𝑥, the derivative of the function. Alternatively, there’s a general result that we can quote. That is, if 𝑓 of 𝑥 is the natural log of some other function 𝑔 of 𝑥, then the derivative 𝑓 prime of 𝑥 is equal to 𝑔 prime of 𝑥 over 𝑔 of 𝑥. Now, we can kind of forget about the three for a moment. We know that the derivative of three times the natural log of two 𝑥 plus four times the natural log of 𝑥 will be equal to three times the derivative of that natural logarithm function. So let’s let 𝑔 of 𝑥 be equal to two 𝑥 plus four times the natural log of 𝑥. Then we know that the derivative 𝑔 prime of 𝑥 is equal to two plus four times one over 𝑥. And that’s because the derivative of the natural log of 𝑥 is simply one over 𝑥.

We can simplify this and we see that 𝑔 prime of 𝑥, the derivative of two 𝑥 plus four times the natural log of 𝑥 is two plus four over 𝑥. 𝑓 prime of 𝑥, the derivative of our function, is therefore three times 𝑔 prime of 𝑥 over 𝑔 of 𝑥. That’s three times two plus four over 𝑥 over two 𝑥 plus four times the natural log of 𝑥. Now, usually we’d look to simplify this expression somewhat. But here we’re being asked to find the value of 𝑓 prime of one. So we substitute 𝑥 equals one into our expression for the derivative. And we get three times two plus four over one over two times one plus four times the natural log of one. Well, we know the natural log of one is zero. So four times the natural log of one is also zero. And we see that 𝑓 prime of one becomes three times six over two which is nine. 𝑓 prime of one is equal to nine. In our next examples, we’ll consider how we can use other rules for differentiation to evaluate the derivative of natural logarithmic functions.

Find d𝑦 by d𝑥, given that 𝑦 equals four times the natural log of 𝑥 plus three over four times the natural log of 𝑥 minus seven.

In this question, we have a fraction or a quotient. This tells us we can use the quotient rule to find the derivative d𝑦 by d𝑥. This says that the derivative of the quotient of two differentiable functions 𝑢 and 𝑣 is 𝑣 time. d𝑢 by d𝑥 minus 𝑢 times d𝑣 by d𝑥 all over 𝑣 squared. We let 𝑢 be equal to four times the natural log of 𝑥 plus three. And 𝑣 is equal to the denominator of our fraction. That’s four times the natural log of 𝑥 minus seven. We then quote the general result for the derivative of the natural log of 𝑥; it’s one over 𝑥. And since the derivative of a constant is zero, we see that d𝑢 by d𝑥 is equal to four lots of one over 𝑥, which is simply four over 𝑥. And similarly d𝑣 by d𝑥 is also four over 𝑥. d𝑦 by d𝑥 is equal to 𝑣 times d𝑢 by d𝑥 minus 𝑢 times d𝑣 by d𝑥 all over 𝑣 squared.

Let’s multiply the numerator and denominator of this fraction by 𝑥 to simplify. When we do, we see that d𝑦 by d𝑥 is equal to four times four times the natural log of 𝑥 minus seven minus four times four times the natural log of 𝑥 plus three all over 𝑥 times four times the natural log of 𝑥 minus seven squared. We distribute the parentheses on our numerator. And we see that we have 16 times the natural log of 𝑥 minus 16 times the natural log of 𝑥 which gives us zero. And we found the derivative of our quotient. It’s negative 40 over 𝑥 times four times the natural log of 𝑥 minus seven squared.

Find the first derivative of the function 𝑦 equals negative seven 𝑥 to the fourth power times the natural log of six 𝑥 to the fourth power.

Here, we have the product of two differentiable functions. The first is negative seven 𝑥 to the fourth power and the second is the natural log of six 𝑥 of the fourth power. We can, therefore, find the first derivative of our function by using the product rule. This says that the derivative of the product of two differentiable functions, 𝑢 and 𝑣, is 𝑢 times d𝑣 by d𝑥 plus 𝑣 times d𝑢 by d𝑥. If we let 𝑢 be equal to negative seven 𝑥 to the fourth power, then d𝑢 by d𝑥 is equal to four times negative seven 𝑥 cubed. That’s negative 28𝑥 cubed. We then let 𝑣 be equal to the natural log of six 𝑥 to the fourth power.

So how do we obtain d𝑣 by d𝑥? Well, we can do one of two things. We could spot that we have a composite function and use the chain rule to find its derivative. Alternatively, we can quote the general result for the derivative of the logarithm of some function 𝑓 of 𝑥. This says that if 𝑦 is equal to the natural log of this function 𝑓 of 𝑥, then d𝑦 by d𝑥 is equal to 𝑓 prime of 𝑥, the derivative of that function, divided by the original function 𝑓 of 𝑥. In this case, 𝑓 of 𝑥 is equal to six 𝑥 to the fourth power. So 𝑓 prime of 𝑥, the derivative of this, is four times six 𝑥 cubed, which is 24𝑥 cubed. d𝑣 by d𝑥 is, therefore, 24𝑥 cubed, divided by the original function, six 𝑥 to the fourth power.

Well, this simplifies quite nicely to four over 𝑥. And we can now substitute everything we have into the formula for the product rule. d𝑦 by d𝑥 is 𝑢 times d𝑣 by d𝑥. That’s negative seven 𝑥 to the fourth power times four over 𝑥 plus 𝑣 times d𝑢 by d𝑥. That’s the natural log of six 𝑥 to the fourth power times negative 28𝑥 cubed. We simplify, and then we factor negative 28𝑥 cubed. And we obtained d𝑦 by d𝑥 to be negative 28𝑥 cubed times the natural logarithm six 𝑥 to the fourth power plus one. In our next example, we’ll look at how we can differentiate functions, which are a combination of both trigonometric and logarithmic functions.

Differentiate 𝑓 of 𝑥 equals five times sin of five times the natural log of 𝑥. This is a composite function. It’s a function of a function. And we can, therefore, use the chain rule to find its derivative. This says that if 𝑦 is some function in 𝑢 and 𝑢 is some function in 𝑥, then d𝑦 by d𝑥 is equal to d𝑦 by d𝑢 times d𝑢 by d𝑥. We’re going to let 𝑢 be equal to five times the natural log of 𝑥. And the derivative of the natural log of 𝑥 is one over 𝑥. So d𝑢 by d𝑥 is five times it; it’s five times one over 𝑥, which is simply five over 𝑥. Instead of using 𝑓 of 𝑥, let’s use 𝑦. And this means that 𝑦 is equal to five sin of 𝑢. And the derivative of sin of 𝑢 is cos of 𝑢. So d𝑦 by d𝑢 is five cos 𝑢. And according to the chain rule d𝑦 by d𝑥 is the product of these.

Referring back to the original notation, we can say that 𝑓 prime of 𝑥, the derivative, is five over 𝑥 times five cos of 𝑢. We’ll replace 𝑢 with five times the natural log of 𝑥. And we see that, in terms of 𝑥, 𝑓 prime of 𝑥 is five over 𝑥 times five cos of five times the natural log of 𝑥. And then we simplify and we see theat 𝑓 prime of 𝑥 is 25 over 𝑥 times the cosine of five times the natural log of 𝑥. In our final example, we’ll consider the higher order derivative of a logarithmic function.

Given that 𝑓 of 𝑥 is equal to nine times the natural log of 𝑥, find the fourth derivative of 𝑓 of 𝑥.

This question is asking us to differentiate our function with respect to 𝑥 four times. So let’s do this and see if we can spot any patterns. We’ll begin by finding the first derivative, 𝑓 prime of 𝑥. We know that the derivative of the natural log of 𝑥 is one over 𝑥, so 𝑓 prime of 𝑥 is nine times one over 𝑥, which is simply nine over 𝑥. We can write this alternatively as nine 𝑥 to the power of negative one, which is going to allow us to find the next derivative. That’s 𝑓 double prime or 𝑓 prime prime of 𝑥. And he We recall the power rule for differentiation. We multiply the expression by the power and then subtract one from the power. And this means 𝑓 double prime of 𝑥 and the derivative of nine 𝑥 to the power of negative one is negative one times nine 𝑥 to the power of negative two. That’s negative nine 𝑥 to the negative two.

We can continue this to find the third derivative, sometimes written as 𝑓 prime prime prime or 𝑓 with a three in brackets here as shown. This is negative two times negative nine 𝑥 to the power of negative three, which is positive 18𝑥 to the negative three. We repeat this process one more time to find the fourth derivative. That’s negative three times 18𝑥 to the power of negative four. And the fourth derivative of nine times the natural log of 𝑥 is negative 54𝑥 to the power of negative four.

And this example demonstrates general result. And that is the 𝑛th order derivative of the natural logarithm function can be expressed as negative one to the power of 𝑛 minus one times 𝑛 minus one factorial all over 𝑥 to the 𝑛th power. This is a nice formula to spot, though it’s also useful to be able to confidently apply the processes by differentiating the natural logarithm function.

In this video, we’ve learned that the derivative of the natural logarithm of 𝑥 is equal to one over 𝑥 when 𝑥 is greater than zero. We’ve seen that if we have the natural logarithm of some function of 𝑥, we can use the chain rule to find the derivative or, alternatively, we can quote the result that d𝑦 by d𝑥 is equal to 𝑓 prime of 𝑥 over 𝑓 of 𝑥. We also saw that these rules can be used in conjunction with the standard rules of differentiation, including the product rule, the quotient rule, and the rules for finding higher order derivatives.

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